I'm trying to immediately invoke a function without using IIFE pattern (enclosing a function definition inside parentheses). Here I see two scenarios:
When a function declaration is invoked immediately: gives
SyntaxError
.When a function expression is invoked immediately: executes successfully.
Example 1: gives SyntaxError
//gives `SyntaxError`
function() {
console.log('Inside the function');
}();
Example 2: Executes without any error
// Executes without any error
var x = function() {console.log('Inside the function')}(); // Inside the function
So, I have these doubts:
- With this approach, why does it give an error for function declaration but not for function expression?
- Is there a way we can immediately invoke a function declaration without using IIFE pattern?
In your code you don't have name for the function that's the reason for syntax error. Even if you would had name it would have thrown error.
function func(){
console.log('x')
}();
The reason is the function declaration doesn't return the values of the function however when you wrap function declaration inside ()
it forces it be a function expression which returns a value.
In the second example the function() {console.log('Inside the function')}
is considered expression because it's on RightHandSide. So it executes without an error.
Is there a way we can immediately invoke a function declaration without using IIFE pattern
You can use +
which will make function declaration an expression.
+function(){
console.log('done')
}()
If you don't want to use +
and ()
you can use new
keyword
new function(){
console.log('done')
}
Extra
A very interesting question is asked by @cat in the comments. I try to answer it.There are three cases
+function(){} //returns NaN
(+function(){return 5})() //VM140:1 Uncaught TypeError: (+(intermediate value)) is not a function
+function(){return 5}() //5
+function(){}
returns NaN
+
acts as Unary Plus here which parses the value next to it to number. As Number(function(){})
returns NaN
so it also returns NaN
(+function(){return 5;})()
returns Error
Usually IIFE are created using ()
. ()
are used to make a function declaration an expression +
is short way for that. Now +function(){}
is already an expression which returns NaN
. So calling NaN
will return error. The code is same as
Number(function(){})()
+function(){return 5;}()
returns 5
In the above line +
is used to make a statement an expression. In the above example first function is called then +
is used on it to convert it to number. So the above line is same as
Number(function(){return 5}())
In the proof of statement "+ runs on after the function is called" Consider the below snippet
console.log(typeof +function(){return '5'}());
So in the above snippet you can see the returned value is string '5'
but is converted to number because of +
A function declaration, like
function foo() {
}
defines (and hoists) the variable name foo
as a function in the current scope. A function declaration doesn't evaluate to a value; it just does something, a bit like an if
does something (rather than evaluate to a value).
You can only invoke values which are functions, eg
<somevalue>()
where somevalue
is a variable name that refers to a function.
Note that function declarations require function names, because otherwise there's no variable name to assign the function to - your original
//gives `SyntaxError`
function() {
console.log('Inside the function');
}();
throws not because of the ()
at the end, but because of the lack of a name.
You can put parentheses at the end of a function declaration, as long as there's something inside the parentheses - but these parentheses do not call the function, they evaluate to a value:
function x() {
console.log('Inside the function');
}(console.log('some expression inside parentheses'));
The interpreter sees these as two separate statements, like
function x() {
console.log('Inside the function');
}
// completely separate:
(console.log('some expression inside parentheses'));
The inside of the parentheses gets evaluated and then discarded, since it doesn't get assigned to anything.
(Empty parentheses are forbidden because they can't be evaluated to a value, similar to how const foo = ()
is forbidden)
The E in IIFE stands for expression, and without the wrapping parenthesis your function is not evaluated as an expression thus the syntax error.
creating an expression is a way of tricking the interpreter and be able to invoke the function immediatly
(function() {
console.log('Inside the function');
})();
In your example you have a function statement followed by the grouping operator, but it's syntactically incorrect for two reasons, first it doesn't have a name, and second because the grouping operator must have an expression inside it, infact if you add a valid one the error will disappear, still you won't obtain your desired result.
function foo() {
console.log('Inside the function');
}();
function foo() {
console.log('Inside the function');
}(1+2);
In order to invoke something, it has to be a function value, a declaration just declares a name, and does not evaluate to the function value itself, hence you cannot invoke it.
A declaration cannot be invoked for the above reason. You have to end up with an expression somehow, either through assignment or grouping (IIFE). So that is a no.
If you give us more context on why you would want to do that, maybe we can help with suggestions.
Not sure why you would want to do it, but:
Is there a way we can immediately invoke a function declaration without using IIFE pattern?
Well, if for function declaration you mean using the keyword function
as in:
function name() { return this.name; }
As far as I know, no. You need the extra parentheses to tell the runtime not to assign the function to a name
variable, if I understand this stuff right.
Now, what you actually can do is to use Function
as in:
new Function('console.log("ey there")')();
Which will execute the console.log
code. No need for IIFE here. But I don't see how this could be better than an IIFE.
you can call in either below ways -
~function(){console.log("hi")}()
!function(){console.log("hi")}()
+function(){console.log("hi")}()
-function(){console.log("hi")}()
(function(){console.log("hi")}());
var i = function(){console.log("hi")}();
true && function(){ console.log("hi") }();
0, function(){ console.log("hi") }();
new function(){ console.log("hi") }
new function(){ console.log("hi") }()
来源:https://stackoverflow.com/questions/56108429/immediately-invoked-function-expression-without-using-grouping-operator